Question

Find a particular solution satisfying the given conditions. $x{y}^{\prime }-y=x^2,y=6$ when $x=2$

Short answer

The particular solution is $$y=x^2+x$$

Step-by-step solution

Rewrite the given differential equation

The given differential equation is $$x{y}^{\text{textquotesingle}}-y=x^2$$Rearrange this to isolate the derivative term $$x\frac{dy}{dx}=y+x^2$$

Separate the variables

Separate the variables to get all terms involving y on one side and all terms involving x on the other:$$x\frac{dy}{dx}-y=x^2$$Divide both sides by x: $$\frac{dy}{dx}=\frac{y}{x}+x$$

Use the method of integrating factors

Identify the standard form of the first-order linear differential equation: $$\frac{dy}{dx}-\frac{y}{x}=x$$The integrating factor, $\mu (x)$, is given by $$\text{Extra close brace or missing open brace}$$

Multiply through by the integrating factor

Multiply both sides of the differential equation by the integrating factor: $$\frac{1}{x}\frac{dy}{dx}-\frac{y}{x^2}=1$$This simplifies to: $$\frac{d}{dx}\left(\frac{y}{x}\right)=1$$

Integrate both sides

Integrate both sides with respect to x: $$\int \frac{d}{dx}\left(\frac{y}{x}\right)dx=\int 1dx$$This yields $$\frac{y}{x}=x+C$$

Solve for y

Solve for y in terms of x: $$y=x^2+Cx$$

Apply the initial condition

Use the initial condition given, $y=6$ when $x=2$, to find C:$$6=(2{)}^2+C(2)$$This simplifies to $$6=4+2C$$Solve for C: $$2=2C$$$$C=1$$

Write the particular solution

Substitute C back into the general solution: $$y=x^2+x$$

Key concepts

Integrating Factors

In differential equations, an integrating factor is used to simplify and solve linear first-order equations, especially when separation of variables is not easily applicable. To find an integrating factor, consider the general form of a linear first-order differential equation written as:

$\frac{dy}{dx}+P(x)y=Q(x)$.
The integrating factor $\mu (x)$ is determined using:
$\mu (x)=e^{\int P(x)dx}$.

In our solved problem, the equation obtained was:
$\frac{dy}{dx}-\frac{y}{x}=x$.
Here, $P(x)=-\frac{1}{x}$, and thus the integrating factor is:
$\mu (x)=e^{-\int \frac{1}{x}dx}=e^{-\ln |x|}=\frac{1}{x}$.

Multiplying the original differential equation by the integrating factor results in:
$\frac{1}{x}\frac{dy}{dx}-\frac{y}{x^2}=1$.
This simplifies our equation, making it easier to solve.

Separation of Variables

Separation of variables is a method to solve differential equations by moving all terms involving one variable to one side of the equation and all terms involving the other variable to the opposite side. However, in linear first-order differential equations, separation of variables may not always be directly usable.

In our example, the equation
$x\frac{dy}{dx}-y=x^2$
was rearranged to get:
$\frac{dy}{dx}=\frac{y}{x}+x$.

Direct separation was not feasible. Hence, we employed integrating factors to handle this equation effectively. It is crucial to recognize when separation of variables can be applied and when other methods, like integrating factors, are more suitable.

Always check and rewrite the differential equation to determine the best approach.

Initial Conditions

Initial conditions are specific values given for the variables in a differential equation, used to find a particular solution. They help in determining the constant of integration when solving differential equations.

For example, the initial condition provided in the problem was $y=6$ when $x=2$.
After solving the differential equation and finding the general solution:
$y=x^2+Cx$,
we used this initial condition to find $C$.

When $x=2$ and $y=6$:
$6=4+2C$
We solved for $C$ and found that $C=1$.

Thus, the particular solution is determined, satisfying the initial condition. It's always crucial to apply initial conditions last, after solving the differential equation to its general form.

Particular Solution

Finding a particular solution involves using the general solution of a differential equation and applying any given initial conditions to find specific values for the constants involved.

In our problem, once the general solution was determined to be$y=x^2+Cx$,
we then used the initial condition $y=6$ when $x=2$. By substituting these values in, we solved for $C$:
$6=4+2C$,
leading to $C=1$.

Substituting $C$ back into the general solution, we obtained the particular solution:
$y=x^2+x$.

The particular solution is essential because it not only solves the differential equation but also fits the given initial conditions, making it unique to the problem at hand.